$2Zn+O_2\to 2ZnO$ $$ $$ $$ $∆G°=-616$ $J$
$2Zn+S_2\to 2ZnS$ $$ $$ $$ $$ $∆G°=-293$ $J$
$S_2+2O_2\to 2S{O}_2$ $$ $$ $$ $$ $∆G°=-408J$

$∆G°$ for the following reaction is:

$2ZnS+3O_2\to 2ZnO+2S{O}_2$

Given that the standard heats of combustion (ΔH°) of C(s), H₂(g), and CH₄(g) are −94 kcal/mol, −68 kcal/mol, and −213 kcal/mol, respectively.
What is the standard enthalpy change (ΔH°) for the reaction:

C(s) + 2H₂(g) → CH₄(g)?

The formation of a solution from two components can be considered as:

The solution so formed will be ideal if:

1. ∆H_Soln = ∆H₁ + ∆H₂ + ∆H₃

2. ∆H_Soln = ∆H₁ + ∆H₂ – ∆H₃

3. ∆H_Soln = ∆H₁ – ∆H₂ – ∆H₃

4. ∆H_Soln = ∆H₃ – ∆H₁ – ∆H₂

For which one of the following equations is $∆\mathrm{H}{°}_{\mathrm{reaction}}$ equal to $∆\mathrm{H}{°}_{\mathrm{formation}}$ for the product:

1. N₂(g) + O₃(g) → N₂O₃(g)

2. CH₄(g) + 2Cl₂(g) → CH₂Cl₂(l) + 2HCl(g)

3. Xe(g) + 2F₂(g) → XeF₄(g)

4. 2CO(g) + O₂(g) → 2CO₂(g)

If the bond energies of H–H, Br – Br, and H –Br are 433, 192, and 364 kJ mol^–1 respectively, the ∆Hº for the reaction H₂(g) + Br₂(g) → 2HBr(g) will be: